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3 years ago

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a train is moving 35m/s to the east. it speeds up to 42 m/s east over the course of 40 seconds. what is the trains acceleration?

1 answer:

lianna [129]3 years ago

7 0

Answer:

0.175 m/s²

Explanation:

Given:

v₀ = 35 m/s

v = 42 m/s

t = 40 s

Find: a

v = at + v₀

42 m/s = a (40 s) + 35 m/s

a = 0.175 m/s²

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Block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bloc

masha68 [24]

Explanation:

Given that,

Mass of the block, m = 300 g = 0.3 kg

Linear spring constant, k = 6.5 N/m

(a) Let T is the period of the block's motion. It is given by :

$T=\dfrac{2\pi}{\omega}$

where

$\omega=\sqrt{\dfrac{k}{m}}$ = angular frequency

Also, $\omega=\sqrt{\dfrac{k}{m}}$

$T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}$

$T=\dfrac{2\pi}{\sqrt{\dfrac{6.5}{0.3}}}$

T = 1.34 seconds

(b) The maximum acceleration of the block is, $a_{max}=2\ m/s^2$

The maximum acceleration is given by :

$a_{max}=\omega^2A$

A is the amplitude of the motion,

$A=\dfrac{a_{max}}{\omega^2}$

$A=\dfrac{a_{max}}{(\sqrt{\dfrac{k}{m}})^2}$

$A=\dfrac{ma_{max}}{k}$

$A=\dfrac{0.3\times 2}{6.5}$

A = 0.09 meters

Hence, this is the required solution.

3 0

3 years ago

I don’t know what this means

kobusy [5.1K]

V stands for Velocity. A stands for Acceleration.
Pretty sure the answer to 3 is D.

8 0

3 years ago

What is the upward force that balances the weight of an object on a surface

Alina [70]

Answer:

The support force/ Normal reaction force

Explanation:

Refer to Newtons third law.

"The third law states that when a force acts on a body due to another body, then an equal and opposite force acts simultaneously on that body"

Since we have a downward force acting on your object, in this case gravity/weight, there is an equal and opposite force given off by the surface the object is on. This is called the support force or the normal reaction force

7 0

3 years ago

Read 2 more answers

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 gram

Bogdan [553]

Complete Question

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?

A His results are accurate but not precise.

B His results are neither accurate nor precise.

C His results are both accurate and precise

D His results are precise but not accurate.

Answer:

Correct option is A

Explanation:

From the question we are told that

The yield of the first test $k = 5.2 \ g$

The yield of the second is $u = 1.3 \ g$

The third yield is $p = 8.5 \ g$

The average yield $A = 5.0 \ g$

The know yield is $A_S = 5.1 \ g$

From the data given we see that

$A_S \ne A$

Since his average yield is closer to the known yield then the answer is accurate

But since the yield for each test are not repeated the answer is not precise

So the answer is accurate but not precise

4 0

3 years ago

Two molecules of lithium are combined with 225 grams of bromine to form two molecules of lithium bromide. If you end up with 690

4vir4ik [10]

The amount of Li present to start the reaction is 55.18g

<u>Explanation:</u>

2Li + Br₂ → 2LiBr

Molecular weight of Br₂ = 159.808 g/mol

Mass of Br₂ present = 225 g

Moles of Br₂ present during the reaction = 225 / 159.808

m = 1.4

Molecular weight of LiBr = 86.845 g/mol

Mass of LiBr formed = 690 g

moles of LiBr produced = 690 / 86.845

m(LiBr) = 7.95

According to the balanced equation, 2 molecules of Li reacts to for 2 molecule of LiBr

So, 7.95 moles of LiBr would require 7.95 moles of Li

The molecular weight of Li is 6.941 g/mol

Thus, the amount of lithium present to start the reaction is

$moles = \frac{given weight}{molecular weight} \\\\7.95 = \frac{w}{6.941} \\\\w = 55.18g$

Therefore, the amount of Li present to start the reaction is 55.18g

6 0

3 years ago